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July 10, 2010 page compiled on July 2, 2015 at 7:05pm

1 Introduction

2 Algorithm diagram

3 Design steps

3.1 backward transformation

3.2 Determine ﬁlter order

3.3 Finding stables poles assuming

3.4 Rescaling the poles

3.5 Converting normalized low pass using frequency transformation

4 Example designs

4.1 Example 1

4.1.1 backward transformation

4.1.2 Determining ﬁlter order

4.2 Finding stables poles assuming

4.2.1 Rescale the poles

4.2.2 Convert normalized low pass using frequency transformation

5 References

2 Algorithm diagram

3 Design steps

3.1 backward transformation

3.2 Determine ﬁlter order

3.3 Finding stables poles assuming

3.4 Rescaling the poles

3.5 Converting normalized low pass using frequency transformation

4 Example designs

4.1 Example 1

4.1.1 backward transformation

4.1.2 Determining ﬁlter order

4.2 Finding stables poles assuming

4.2.1 Rescale the poles

4.2.2 Convert normalized low pass using frequency transformation

5 References

This is a detailed review of low pass Butterworth analog ﬁlter design. The goal is to generate Butterworth transfer function from frequency speciﬁcations. The following are the four speciﬁcations of the design

The passband corner frequency in Hz | |

The stopband corner frequency in Hz | |

The attenuation in db at | |

The attenuation in db at | |

This diagram below illustrates these speciﬁcations

The speciﬁcations are given in db (the left diagram above) and not in magnitude (right diagram above).

The speciﬁcations are given with reference to the transfer function magnitude. The phase is not taken into account in the speciﬁcations. Butterworth analog transfer function transfer function magnitude is given by

Where is the cutoﬀ frequency. This is the frequency at which

The goal of the design is to determine and from the speciﬁcations. Once and are found, poles are found. Once the poles are found, then is now determined.

The following diagram outlines the design algorithm

Solving for from the above gives

| (1) |

Using gives

Solving for from the above results in

| (2) |

Substituting found in (1) into the above gives

| (3) |

Solving for in the above by taking logs gives

| (4) |

Since the order of ﬁlter is an integer, the above value is rounded upwards to the next integer if it is not an integer. Let this new be to make it clear that this is an updated from the original .

Since Butterworth magnitude square of the transfer function is

Hence poles are found by setting the denominator of the above to zero. Setting gives

Only the LHS poles are needed, which are located at , because these are the stable poles.

Now thatthe poles are found, becomes

| (6) |

Either or have to be adjusted depending on if the excess tolerance is to be assigned to the passband or to the stop band and is calculated based on this.

If the excess tolerance is to be assigned to the passband, then (3) is solved for and this new found value is called

| (6.1) |

Also needs to be determined from (1). Calling this to reﬂect that this goes with the updated and not the original

| (6.2) |

However, if the excess tolerance is to be assigned to the stopband, then (3) is solved for and this new found value is called

| (6.3) |

is adjusted to .

From (2), and using the above new value of gives

| (6.4) |

found above in (6) is now adjusted since that was found for and now an updated is found. To do that is replaced by , hence becomes

The ﬁrst part of the design is now complete. is found and adjusted or depending on the requirements for excess tolerance. All the parts needed are found to design by ﬁnding its poles. Adjusted values must be used from now on.

The above found in (7) was designed for frequency and . The above is called the normalized transfer function. It is a low pass analog ﬁlter, which needs to be mapped to a low pass analog ﬁlter, but un-normalized based on the actual frequencies speciﬁed (Since the above was designed based on using ).

Adjustment is now now made to obtain for and .

To do this, above is replaced by . Equation (7) becomes

The zeros of are located at and there are of them.

When simplifying the denominator above, the complex conjugate terms are multiplied with each others to obtain real coeﬃcients.

Given

hz, hz, db, db, and Excess tolerance at stopband, determine

From (4)

Hence

From (5), and since

Find the poles

Excess tolerance is in the stopband, hence from (6.3)

Hence new is found from (6.4)

Hence the above becomes (using equation 7 as reference)

replace by , hence becomes (using equation 8 as reference), and noting that

Now multiplying the complex conjugate terms with each others (to remove the complex terms) gives

- 1.
- ECE 408 lecture notes chapter 12, by Dr James S. Kang. Cal Poly pomona, California, USA.
- 2.
- Mostafa Shiva, Electrical engineering department, California state university, Fullerton, Lecture notes, handout H.
- 3.
- John Proakis, Dimitris Manolakis, digital signal processing, 3rd edition